Question

Evaluate the Limit, if it exists

Lim (1/4 + 1/x)/(4+x)
X-->-4

Answer 1

Use the L'Hôpital's rule:
\[\lim_{x \rightarrow -4} \frac{\frac{1}{4} + \frac{1}{x}}{4 + x} =\lim_{x \rightarrow -4}\frac{f(x)}{g(x)} = \lim_{x \rightarrow -4}\frac{f'(x)}{g'(x)} = \lim_{x \rightarrow -4} \frac{-\frac{1}{x^2}}{1} = -\frac{1}{16}\]

Answer 2

how did you get (-1/x^2)/1

Answer 3

\[\lim_{x \to -4} \frac{\frac{1}{x} + \frac{1}{4}}{4+x} \implies \lim_{x \to -4} \frac{\frac{4+x}{4x}}{4+x} \implies \lim_{x \to -4} \frac{1}{4x} \implies \frac{1}{-16}\]Without L'Hospital...

Answer 4

(-1/x^2) is the derivative of 1/4 + 1/x, and 1 of 4 + x